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Difference between revisions of "2000 AMC 12 Problems/Problem 8"

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<math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math>
+
<math>\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801</math>
  
−
==Solution==
+
==Video:==
 +
https://www.youtube.com/watch?v=HVP6qjKAkjA&t=2s
  
−
===Solution 1===
+
== Solution 1 ==
 +
We can attempt <math>0^2+1^2=1</math> and <math>1^2+2^2=5</math>, so the pattern here looks like the number of squares in the <math>n</math>-th figure is <math>n^2+(n+1)^2</math>. When we plug in 100 for <math>n</math>, we get <math>100^2+101^2=10000+10201=20201</math>, or option <math>\textbf{(C)}</math>.
  
−
We have a recursion:
+
==Solution 2==
−
 
 
−
<math>A_n=A_{n-1}+4n</math>.
 
−
 
 
−
We add increasing multiples of <math>4</math> each time we go up a figure. So, for example, in figure <math>3</math>, we would have <math>4 \cdot 1+4 \cdot 2+4 \cdot 3</math> added on to Figure 0, which is <math>1</math>. This yields <math>4+8+12+1 = 25.</math>
 
−
 
 
−
So, to go from Figure 0 to 100, we add
 
−
 
 
−
<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>.
 
−
 
 
−
We then add <math>20200</math> to the number of squares in Figure 0 to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math>
 
−
 
 
−
===Solution 2===
 
−
 
 
−
We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>:
 
−
 
 
−
<asy>
 
−
draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);
 
−
draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);
 
−
draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);
 
−
draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);
 
−
draw((3,0)--(3,7));
 
−
</asy>
 
−
 
 
−
The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math>
 
−
 
 
−
===Solution 3===
 
  
 
Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>.  Thus, the original sequence can be generated from a quadratic function.
 
Using the recursion from solution 1, we see that the first differences of <math>4, 8, 12, ...</math> form an arithmetic progression, and consequently that the second differences are constant and all equal to <math>4</math>.  Thus, the original sequence can be generated from a quadratic function.
Line 87: Line 63:
 
<math>f(n) = 2n^2 + 2n + 1</math>
 
<math>f(n) = 2n^2 + 2n + 1</math>
  
−
Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\text{C}}</math>
+
Calculating the answer to our problem, <math>f(100) = 20000 + 200 + 1 = 20201</math>, which is choice <math>\boxed{\textbf{(C) }20201}</math>.
 +
 
 +
==Solution 3==
 +
We can see that each figure <math>n</math> has a central box and 4 columns of <math>n</math> boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are <math>\sum_{n=1}^{100} n = 5050</math> squares. <math>4 \cdot 5050 = 20200</math>. Adding in the original center box we have <math> 20200 + 1 = \boxed{\textbf{(C) }20201}</math>.
 +
 
 +
==Solution 4==
 +
Let <math>a_n</math> be the number of squares in figure <math>n</math>.  We can easily see that
 +
<cmath>a_0=4\cdot 0+1</cmath>
 +
<cmath>a_1=4\cdot 1+1</cmath>
 +
<cmath>a_2=4\cdot 3+1</cmath>
 +
<cmath>a_3=4\cdot 6+1.</cmath>
 +
See that we multiply the number we are on to the next consecutive number.
 +
Note that in <math>a_n</math>, the number multiplied by the 4 is the <math>n</math>th triangular number.  Hence, <math>a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}</math>.
 +
~ edited by mathlover66
 +
 
 +
==Solution 5==
 +
Let <math>f_n</math> denote the number of unit cubes in a figure. We have
 +
<cmath>f_0=1</cmath>
 +
<cmath>f_1=5</cmath>
 +
<cmath>f_2=13</cmath>
 +
<cmath>f_3=25</cmath>
 +
<cmath>f_4=41</cmath>
 +
<cmath>...</cmath>
 +
 
 +
Computing the difference between the number of cubes in each figure yields
 +
<cmath>4,8,12,16,...</cmath>
 +
It is easy to notice that this is an arithmetic sequence, with the first term being <math>4</math> and the difference being <math>4</math>. Let this sequence be <math>a_n</math>
 +
 
 +
From <math>f_0</math> to <math>f_{100}</math>, the sequence will have <math>100</math> terms. Using the arithmetic sum formula yields
 +
 
 +
<cmath>S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}</cmath>
 +
<cmath>=50(2\cdot 4+99\cdot 4)</cmath>
 +
<cmath>=50(101\cdot 4)</cmath>
 +
<cmath>=200\cdot 101</cmath>
 +
<cmath>=20200</cmath>
 +
 
 +
So <math>f_{100}=1+20200=\boxed{\textbf{(C) }20201}</math> unit cubes.
 +
 
 +
~ljlbox
 +
 
 +
== Solution 6 (Newton's Forward Differences) ==
 +
We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) <math>\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201</math> or <math>\textbf{(C)}</math>
 +
-vsamc
 +
 
 +
 
 +
== Solution 7 (Newton’s little formula) ==
 +
Newton’s little formula states that <math>a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}</math> if first term is <math>a_1</math> and <math>A =</math> first difference, <math>B =</math> second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is <math>a_{101}</math>): <math>a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201} = \textbf{(C)}.</math>
 +
~Peelybonehead
 +
~clarification by LeonidasTheConquerer
 +
 
 +
== Solution 8 (geometrical intuition) ==
 +
By taking figure <math>n</math>, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference):
 +
<asy>
 +
import olympiad;
  
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===Solution 4===
+
int fig = 3;
−
We can see that each figure <math>n</math> has a central box and 4 columns of <math>n</math> boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are <math>\sum_{n=1}^{100} n = 5050</math> squares in each quarter. <math>4 \cdot 5050 = 20200</math>. Adding in the original center box we have <math> 20200 + 1 \implies \boxed{\text{C}}</math>
+
 
 +
for (int i = 0; i <= 2*fig; ++i) {
 +
  for (int j = 0; j <= 2*fig; ++j) {
 +
    pair p = (i, j);
 +
    //do we need this point?
 +
    if ((i+j)%2 == 1) {
 +
      continue;
 +
    }
 +
    //draw squares
 +
    draw((p+N)--(p+E)--(p+S)--(p+W)--cycle);
 +
    //draw lattices
 +
    if (i < 2*fig-1) {
 +
      draw(p--p+2*E, red);
 +
    }
 +
    if (j < 2*fig-1) {
 +
      draw(p--p+2*N, red);
 +
    }
 +
    //mark center points
 +
    if (i % 2 == 0) {
 +
      dot(p, green);
 +
    } else {
 +
      dot(p, blue);
 +
    }
 +
  }
 +
}
 +
</asy>
 +
There are two lattices of points (shown in red): one of side length <math>n</math> (blue) and one of <math>n+1</math> (green). The rest follows like in solution 1.
 +
 
 +
-[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:21, 29 July 2023 (EDT)
 +
 
 +
== Solution 9 ==
 +
If you do not want to base off the previous number in the sequence, you can come up with a formula based on which figure you are looking for:
 +
 
 +
<math>1+x(x-2+x)</math>
 +
 
 +
*However, this formula says that in figure 1, there is 1 square and so forth. Therefore, if we are looking to find <math>f(100)</math>, we will have to find <math>f(101)</math>.  
 +
 
 +
This is the formula because to get from 1 to 5, we have +4. To get 1 to 13, it is +12. To get 1 to 25, it is +24. Assuming that 5 squares is figure 2 rather than in figure one (sorry, it just made better sense in my brain), then in figure two, it is 2*2 to get +4 on top of one. In figure “3”, to get +12, you have 3*4. In figure 4, to get +24, you have 4*6. Now, you can notice that in figure two, it is 2* (the distance to 2- which is 0- added onto 2) which is +2*2. In figure 3, to get +12, it is 3 (the figure number) * (the distance from 3 to 2- which is 1, added back onto the figure number). In figure 4, to get +24, you can get 4*6 (the figure number * (the distance from 4 to two- which is 2- added back onto the figure number, 6).
 +
 
 +
So to get figure 100, we have to do f(101) because I said that figure 1 was 1 rather than figure 0. So 1+ 101(101-2+101) which is 20201.
 +
 
 +
-Smartgrowth
 +
 
 +
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/y7WrXbKEh18?si=T5jENfI9Wa9UJLuo
 +
 
 +
~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==
Line 96: Line 172:
 
{{AMC12 box|year=2000|num-b=7|num-a=9}}
 
{{AMC12 box|year=2000|num-b=7|num-a=9}}
 
{{AMC10 box|year=2000|num-b=11|num-a=13}}
 
{{AMC10 box|year=2000|num-b=11|num-a=13}}
−
 
+
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:15, 25 July 2024

The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.

Problem

Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]


$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Video:

https://www.youtube.com/watch?v=HVP6qjKAkjA&t=2s

Solution 1

We can attempt $0^2+1^2=1$ and $1^2+2^2=5$, so the pattern here looks like the number of squares in the $n$-th figure is $n^2+(n+1)^2$. When we plug in 100 for $n$, we get $100^2+101^2=10000+10201=20201$, or option $\textbf{(C)}$.

Solution 2

Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can be generated from a quadratic function.

If $f(n) = an^2 + bn + c$, and $f(0) = 1$, $f(1) = 5$, and $f(2) = 13$, we get a system of three equations in three variables:

$f(0) = 1$ gives $c = 1$

$f(1) = 5$ gives $a + b + c = 5$

$f(2) = 13$ gives $4a + 2b + c = 13$

Plugging in $c=1$ into the last two equations gives

$a + b = 4$

$4a + 2b = 12$

Dividing the second equation by 2 gives the system:

$a + b = 4$

$2a + b = 6$

Subtracting the first equation from the second gives $a = 2$, and hence $b = 2$. Thus, our quadratic function is:

$f(n) = 2n^2 + 2n + 1$

Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$, which is choice $\boxed{\textbf{(C) }20201}$.

Solution 3

We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$. Adding in the original center box we have $20200 + 1 = \boxed{\textbf{(C) }20201}$.

Solution 4

Let $a_n$ be the number of squares in figure $n$. We can easily see that \[a_0=4\cdot 0+1\] \[a_1=4\cdot 1+1\] \[a_2=4\cdot 3+1\] \[a_3=4\cdot 6+1.\] See that we multiply the number we are on to the next consecutive number. Note that in $a_n$, the number multiplied by the 4 is the $n$th triangular number. Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}$. ~ edited by mathlover66

Solution 5

Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]

Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$. Let this sequence be $a_n$

From $f_0$ to $f_{100}$, the sequence will have $100$ terms. Using the arithmetic sum formula yields

\[S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}\] \[=50(2\cdot 4+99\cdot 4)\] \[=50(101\cdot 4)\] \[=200\cdot 101\] \[=20200\]

So $f_{100}=1+20200=\boxed{\textbf{(C) }20201}$ unit cubes.

~ljlbox

Solution 6 (Newton's Forward Differences)

We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) $\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201$ or $\textbf{(C)}$ -vsamc


Solution 7 (Newton’s little formula)

Newton’s little formula states that $a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}$ if first term is $a_1$ and $A =$ first difference, $B =$ second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is $a_{101}$): $a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201} = \textbf{(C)}.$ ~Peelybonehead ~clarification by LeonidasTheConquerer

Solution 8 (geometrical intuition)

By taking figure $n$, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference): [asy] import olympiad; int fig = 3; for (int i = 0; i <= 2*fig; ++i) { for (int j = 0; j <= 2*fig; ++j) { pair p = (i, j); //do we need this point? if ((i+j)%2 == 1) { continue; } //draw squares draw((p+N)--(p+E)--(p+S)--(p+W)--cycle); //draw lattices if (i < 2*fig-1) { draw(p--p+2*E, red); } if (j < 2*fig-1) { draw(p--p+2*N, red); } //mark center points if (i % 2 == 0) { dot(p, green); } else { dot(p, blue); } } } [/asy] There are two lattices of points (shown in red): one of side length $n$ (blue) and one of $n+1$ (green). The rest follows like in solution 1.

-Integralarefun (talk) 18:21, 29 July 2023 (EDT)

Solution 9

If you do not want to base off the previous number in the sequence, you can come up with a formula based on which figure you are looking for:

$1+x(x-2+x)$

  • However, this formula says that in figure 1, there is 1 square and so forth. Therefore, if we are looking to find $f(100)$, we will have to find $f(101)$.

This is the formula because to get from 1 to 5, we have +4. To get 1 to 13, it is +12. To get 1 to 25, it is +24. Assuming that 5 squares is figure 2 rather than in figure one (sorry, it just made better sense in my brain), then in figure two, it is 2*2 to get +4 on top of one. In figure “3”, to get +12, you have 3*4. In figure 4, to get +24, you have 4*6. Now, you can notice that in figure two, it is 2* (the distance to 2- which is 0- added onto 2) which is +2*2. In figure 3, to get +12, it is 3 (the figure number) * (the distance from 3 to 2- which is 1, added back onto the figure number). In figure 4, to get +24, you can get 4*6 (the figure number * (the distance from 4 to two- which is 2- added back onto the figure number, 6).

So to get figure 100, we have to do f(101) because I said that figure 1 was 1 rather than figure 0. So 1+ 101(101-2+101) which is 20201.

-Smartgrowth

Video Solution by Daily Dose of Math

https://youtu.be/y7WrXbKEh18?si=T5jENfI9Wa9UJLuo

~Thesmartgreekmathdude

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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